A long solenoid has 800 turns per metre length of solenoid A currentof 1.6 A flow through it. The magnetic induction at the end of the solenoid on its axis is

Options

A. `16xx10^(-4)` Tesla
B. `8xx10^(-4)` Tesla
C. `32xx10^(-4)` Tesla
D. `4xx10^(-4)` Tesla

Answer

To find the magnetic induction (magnetic field) at the end of a long solenoid on its axis, we can follow these steps:

### Step 1: Identify the given parameters
- Number of turns per meter (n) = 800 turns/m
- Current (I) = 1.6 A

### Step 2: Recall the formula for magnetic field inside a solenoid
The magnetic field (B) inside a long solenoid is given by the formula:
\[ B = \mu_0 n I \]
where:
- \( \mu_0 \) = permeability of free space = \( 4\pi \times 10^{-7} \, \text{T m/A} \)
- n = number of turns per unit length
- I = current flowing through the solenoid

### Step 3: Calculate the magnetic field at the center of the solenoid
Using the formula:
\[ B_{\text{center}} = \mu_0 n I \]
Substituting the values:
\[ B_{\text{center}} = (4\pi \times 10^{-7}) \times (800) \times (1.6) \]

### Step 4: Calculate the magnetic field at the end of the solenoid
The magnetic field at the end of the solenoid is given by:
\[ B_{\text{end}} = \frac{B_{\text{center}}}{2} \]
Thus,
\[ B_{\text{end}} = \frac{\mu_0 n I}{2} \]

### Step 5: Substitute the values into the equation
Now substituting the values into the equation for \( B_{\text{end}} \):
\[ B_{\text{end}} = \frac{(4\pi \times 10^{-7}) \times (800) \times (1.6)}{2} \]

### Step 6: Calculate the final result
Calculating this step-by-step:
1. Calculate \( 4\pi \times 10^{-7} \):
\[ 4\pi \approx 12.5664 \]
\[ 4\pi \times 10^{-7} \approx 12.5664 \times 10^{-7} \]
2. Multiply by 800:
\[ 12.5664 \times 10^{-7} \times 800 \approx 10.0531 \times 10^{-4} \]
3. Multiply by 1.6:
\[ 10.0531 \times 10^{-4} \times 1.6 \approx 16.08496 \times 10^{-4} \]
4. Divide by 2:
\[ \frac{16.08496 \times 10^{-4}}{2} \approx 8.04248 \times 10^{-4} \]

Thus, the magnetic induction at the end of the solenoid is approximately:
\[ B_{\text{end}} \approx 8.0 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnetic induction at the end of the solenoid on its axis is \( 8.0 \times 10^{-4} \, \text{T} \).

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